What is expand-(e^{4x}(2e^{-2x}-e^{-x}))/6 ?

The solution to expand-(e^{4x}(2e^{-2x}-e^{-x}))/6 is -(e^{2x})/3+(e^{3x})/6

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expand-(e^{4x}(2e^{-2x}-e^{-x}))/6

expand

- \frac{e ^{4 x} ( 2 e ^{- 2 x} - e ^{- x} )}{6}

- \frac{e ^{2 x}}{3} + \frac{e ^{3 x}}{6}

Solution steps

- \frac{e ^{4 x} ( 2 e ^{- 2 x} - e ^{- x} )}{6}

Expand

e^{4 x} (2 e^{- 2 x} - e^{- x}) : 2 e^{2 x} - e^{3 x}

= - \frac{2 e ^{2 x} - e ^{3 x}}{6}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{2 e ^{2 x} - e ^{3 x}}{6} = - (\frac{2 e ^{2 x}}{6}) - (- \frac{e ^{3 x}}{6})

= - (\frac{2 e ^{2 x}}{6}) - (- \frac{e ^{3 x}}{6})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{2 e ^{2 x}}{6} + \frac{e ^{3 x}}{6}

Cancel

\frac{2 e ^{2 x}}{6} : \frac{e ^{2 x}}{3}

= - \frac{e ^{2 x}}{3} + \frac{e ^{3 x}}{6}

(8 t^{7}, 5 t^{3}, - 4 t^{3}) l e f t (1, 2 t, 3 t^{2})

\frac{4 ( 6 x ^{2} - 3 )}{( 2 x ^{3} + 3 x ) ^{2}}

(1 + x) d x

5 (1)^{2} + 8 (1) z - 44 (1) + 5 z^{2} - 46 z + 104

(1 - \frac{y}{x})^{2}

What is expand-(e^{4x}(2e^{-2x}-e^{-x}))/6 ?
The solution to expand-(e^{4x}(2e^{-2x}-e^{-x}))/6 is -(e^{2x})/3+(e^{3x})/6